A micrometeor has a mass of 0.005 grams. When it enters Earth’s atmosphere, it travels at 21,000 meters per second. What is its kinetic energy when it enters Earth’s atmosphere? KE=12mv2 (1 point) Responses 1,102,500 J 1,102,500 J 2,205 J 2,205 J 0.0525 J 0.0525 J 1,102.5 J

Answer 1

To calculate the kinetic energy (KE) of the micrometeor, we can use the formula:

\[ KE = \frac{1}{2}mv^2 \]

Where:

• \( m \) is the mass in kilograms
• \( v \) is the velocity in meters per second

First, we need to convert the mass from grams to kilograms:

\[ m = 0.005 \text{ grams} = 0.005 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.000005 \text{ kg} \]

Now we can substitute the values into the kinetic energy formula. The velocity \( v \) is already given as 21,000 m/s.

\[ KE = \frac{1}{2} \times 0.000005 \text{ kg} \times (21000 \text{ m/s})^2 \]

Calculating \( (21000 \text{ m/s})^2 \):

\[ (21000)^2 = 441000000 \]

Now substitute this back into the equation:

\[ KE = \frac{1}{2} \times 0.000005 \text{ kg} \times 441000000 \]

Calculating further:

\[ KE = 0.0000025 \times 441000000 = 1102.5 \text{ J} \]

Therefore, the kinetic energy when it enters Earth’s atmosphere is:

\[ \boxed{1102.5 \text{ J}} \]